Chapter 5. Solving Linear Equations and Inequalities
5.1. Objectives*
After completing this chapter, you should
Solving Equations (Section 5.2)
be able to identify various types of equations
understand the meaning of solutions and equivalent equations
be able to solve equations of the form x + a = b and x – a = b .
be familiar with and able to solve literal equation
Solving Equations of the Form
ax
=
b
and 
(Section 5.3)
understand the equality property of addition and multiplication
be able to solve equations of the form ax = b and
Further Techniques in Equation Solving (Section 5.4)
be comfortable with combining techniques in equation solving
be able to recognize identities and contradictions
Applications I - Translating from Verbal to Mathematical Expressions (Section 5.5)
be able to translate from verbal to mathematical expressions
Linear Inequalities in One Variable (Section 5.7)
understand the meaning of inequalities
be able to recognize linear inequalities
know, and be able to work with, the algebra of linear inequalities and with compound inequalities
Linear Inequalities in Two Variables (Section 5.8)
be able to identify the solution of a linear equation in two variables
know that solutions to linear equations in two variables can be written as ordered pairs
5.2. Solving Equations *
Overview
Types of Equations
Solutions and Equivalent Equations
Literal Equations
Solving Equations of the Form x + a = b and x − a = b
Types of Equations
Some equations are always true. These equations are called identities. Identities are equations that are true for all acceptable values of the variable, that is, for all values in the domain of the equation. 5x = 5x is true for all acceptable values of x . y + 1 = y + 1 is true for all acceptable values of y .2 + 5 = 7 is true, and no substitutions are necessary.
Some equations are never true. These equations are called contradictions. Contradictions are equations that are never true regardless of the value substituted for the variable. x = x + 1 is never true for any acceptable value of x . 0 · k = 14 is never true for any acceptable value of k . 2 = 1 is never true.
The truth of some equations is conditional upon the value chosen for the variable. Such equations are called conditional equations. Conditional equations are equations that are true for at least one replacement of the variable and false for at least one replacement of the variable. x + 6 = 11 is true only on the condition that x = 5. y − 7 = − 1 is true only on the condition that y = 6 .
Solutions and Equivalent Equations
The collection of values that make an equation true are called solutions of the equation. An equation is solved when all its solutions have been found.
Some equations have precisely the same collection of solutions. Such equations are called equivalent equations. The equations
are equivalent equations because the only value that makes each one true is 3.
Sample Set A
Tell why each equation is an identity, a contradiction, or conditional.
Example 5.1.
The equation x − 4 = 6 is a conditional equation since it will be true only on the condition that x = 10.
Example 5.2.
The equation x − 2 = x − 2 is an identity since it is true for all values of x . For example,
Example 5.3.
The equation a + 5 = a + 1 is a contradiction since every value of a produces a false statement. For example,
Practice Set A
For each of the following equations, write "identity," "contradiction," or "conditional." If you can, find the solution by making an educated guess based on your knowledge of arithmetic.
Literal Equations
Some equations involve more than one variable. Such equations are called literal equations.
An equation is solved for a particular variable if that variable alone equals an expression that does not contain that particular variable.
The following equations are examples of literal equations.
y = 2x + 7 . It is solved for y .
d = r t . It is solved for d .
I = p r t . It is solved for I .
. It is solved for
z
.y + 1 = x + 4 . This equation is not solved for any particular variable since no variable is isolated.
Solving Equation of the form x + a = b and x − a = b
Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side.
This suggests the following procedures:
We can obtain an equivalent equation (an equation having the same solutions as the original equation) by adding the same number to both sides of the equation.
We can obtain an equivalent equation by subtracting the same number from both sides of the equation.
We can use these results to isolate x , thus solving for x .
Example 5.4. Solving x + a = b for x
Example 5.5. Solving x − a = b for x
Example 5.6. Method for Solving x + a = b and x − a = b for x
To solve the equation x + a = b for x , subtract a from both sides of the equation. To solve the equation x − a = b for x , add a to both sides of the equation.
Sample Set B
Example 5.7.
Solve x + 7 = 10 for x .
Check: Substitute 3 for
x
in the original equation. 
Example 5.8.
Solve m − 2 = − 9 for m .
Check: Substitute
− 7
for
m
in the original equation.
Example 5.9.
Solve
y − 2.181 = − 16.915
for
y
.
On the Calculator
Example 5.10.
Solve y + m = s for y .
Check: Substitute
s − m
for
y
in the original equation.
Example 5.11.
Solve k − 3h = − 8h + 5 for k .
Practice Set B
Exercises
For the following problems, classify each of the equations as an identity, contradiction, or conditional equation.
Exercise 5.2.19.
y − 8 = − 12
Exercise 5.2.21.
k − 2 = k − 3
Exercise 5.2.23.
x + 1 = 0
For the following problems, determine which of the literal equations have been solved for a variable. Write "solved" or "not solved."
Exercise 5.2.25.
m = 2k + n − 1
Exercise 5.2.27.
h k = 2k + h
Exercise 5.2.29.
5m = 2m − 7
For the following problems, solve each of the conditional equations.
Exercise 5.2.31.
h − 8 = 14
Exercise 5.2.33.
m − 2 = 5
Exercise 5.2.35.
y − 8 = − 1
Exercise 5.2.37.
m − 12 = 0
Exercise 5.2.39.
h − 265 = − 547
Exercise 5.2.41.
h − 4.82 = − 3.56
Exercise 5.2.43.
k + 1.0135 = – 6.0032
Exercise 5.2.45.
Solve P + 3Q − 8 = 0 for P.
Exercise 5.2.47.
Solve x − 3y + 5z + 1 = 2y − 7z + 8 for x.
Exercises for Review
Exercise 5.2.51.
(Section 4.2) Write the number of terms that appear in the expression 5x 2 + 2x − 6 + (a + b) , and then list them.
Solutions to Exercises
5.3. Solving Equations of the Form ax=b and x/a=b *
Overview
Equality Property of Division and Multiplication
Solving a x = b and
for
x
Equality Property of Division and Multiplication
Recalling that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side suggests the equality property of division and multiplication, which states:
We can obtain an equivalent equation by dividing both sides of the equation by the same nonzero number, that is, if c ≠ 0, then a = b is equivalent to
.We can obtain an equivalent equation by multiplying both sides of the equation by the same nonzero number, that is, if c ≠ 0, then a = b is equivalent to a c = b c .
We can use these results to isolate x, thus solving the equation for x .
Example 5.12.
Solving a x = b for x
Example 5.13.
Solving 
for
x
Solving
a
x = b
and
for
x
Example 5.14.
Method for Solving 
To solve
a
x = b
for
x
, divide both sides of the equation by
a
. To solve 
for
x
, multiply both sides of the equation by
a
.
Sample Set A
Example 5.15.
Solve 5x = 35 for x .
Example 5.16.
Solve
for
x
.
Example 5.17.
Solve 
for
y
.
Method (1) (Use of cancelling):
Method (2) (Use of reciprocals):
Example 5.18.
Solve the literal equation 
for
x
.
Practice Set A
Round the result to two decimal places.
Exercises
In the following problems, solve each of the conditional equations.
Exercise 5.3.10.
5y = 75
Exercise 5.3.12.
8x = 56
Exercise 5.3.14.
3x = 93
Exercise 5.3.16.
9m = − 108
Exercise 5.3.18.
12q = − 180
Exercise 5.3.20.
− 20x = 100
Exercise 5.3.22.
− 8m = − 40
Exercise 5.3.24.
− 9y = 126
Exercise 5.3.26.
Exercise 5.3.28.
Exercise 5.3.30.
Exercise 5.3.32.
Exercise 5.3.34.
3.06m = 12.546
Exercise 5.3.36.
Exercise 5.3.38.
Exercise 5.3.40.
Exercise 5.3.42.
Exercise 5.3.44.
Solve m 2 n = 2s for n .
Exercise 5.3.46.
Solve 
for
p
.
Exercise 5.3.48.
Solve 
for
pc
.
Exercise 5.3.50.
Solve 
for
□
.
Exercises for Review
Exercise 5.3.53. (Go to Solution)
(Section 4.4) Classify 10x 3 − 7x as a monomial, binomial, or trinomial. State its degree and write the numerical coefficient of each item.
Solutions to Exercises
5.4. Further Techniques in Equation Solving *
Overview
Combining Techniques in Equation Solving
Recognizing Identities and Contrdictions
Combining Techniques in Equation Solving
In Sections Section 5.2 and Section 5.3 we worked with techniques that involved the use of addition, subtraction, multiplication, and division to solve equations. We can combine these techniques to solve more complicated equations. To do so, it is helpful to recall that an equation is solved for a particular variable when all other numbers and/or letters have been disassociated from it and it is alone on one side of the equal sign. We will also note that
To associate numbers and letters we use the order of operations.
Multiply/divide
Add/subtract
To undo an association between numbers and letters we use the order of operations in reverse.
Add/subtract
Multiply/divide
Sample Set A
Example 5.19.
Solve 4x − 7 = 9 for x.
Example 5.20.
Solve 
Example 5.21.
Solve 
for
a.
Practice Set A
Sometimes when solving an equation it is necessary to simplify the expressions composing it.
Sample Set B
Example 5.22.
Solve 4x + 1 − 3x = ( − 2)(4) for x.
Example 5.23.
Solve 3(m − 6) − 2m = − 4 + 1 for m.
Practice Set B
Solve and check each equation.
Often the variable we wish to solve for will appear on both sides of the equal sign. We can isolate the variable on either the left or right side of the equation by using the techniques of Sections Section 5.2 and Section 5.3.
Sample Set C
Example 5.24.
Solve 6x − 4 = 2x + 8 for x.
Example 5.25.
Solve 6(1 − 3x) + 1 = 2x − [3(x − 7) − 20] for x.
Practice Set C
Recognizing Identities and Contradictions
As we noted in Section Section 5.2, some equations are identities and some are contradictions. As the problems of Sample Set D will suggest,
Recognizing an Identity
If, when solving an equation, all the variables are eliminated and a true statement results, the equation is an identity.
Recognizing a Contradiction
If, when solving an equation, all the variables are eliminated and a false statement results, the equation is a contradiction.
Sample Set D
Example 5.26.
Solve 9x + 3(4 − 3x) = 12 for x.
The variable has been eliminated and the result is a true statement. The original equation is an identity.
Example 5.27.
Solve − 2(10 − 2y) − 4y + 1 = − 18 for y.
The variable has been eliminated and the result is a false statement. The original equation is a contradiction.
Practice Set D
Classify each equation as an identity or a contradiction.
Exercises
For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction.
Exercise 5.4.17.
6y − 4 = 20
Exercise 5.4.19.
3x + 4 = 40
Exercise 5.4.21.
8k − 7 = − 23
Exercise 5.4.23.
7a + 2 = − 26
Exercise 5.4.25.
14x + 1 = − 55
Exercise 5.4.27.
Exercise 5.4.29.
Exercise 5.4.31.
Exercise 5.4.33.
Exercise 5.4.35.
Exercise 5.4.37.
3(x − 6) + 5 = − 25
Exercise 5.4.39.
6x + 14 = 5x − 12
Exercise 5.4.41.
− 3m + 1 = 3m − 5
Exercise 5.4.43.
12n + 5 = 5n − 16
Exercise 5.4.45.
− 4(5y + 3) + 5(1 + 4y) = 0
Exercise 5.4.47.
4(4y + 2) = 3y + 2[1 − 3(1 − 2y)]
Exercise 5.4.49.
12 − (m − 2) = 2m + 3m − 2m + 3(5 − 3m)
Exercise 5.4.51.
3[4 − 2(y + 2)] = 2y − 4[1 + 2(1 + y)]
For the following problems, solve the literal equations for the indicated variable. When directed, find the value of that variable for the given values of the other variables.
Exercise 5.4.53.
Solve 
for
R. Find the value of
R
when
I = 0.005 and
E = 0.0035.
Exercise 5.4.54. (Go to Solution)
Solve P = R − C for R. Find the value of R when P = 27 and C = 85.
Exercise 5.4.55.
Solve 
for
x. Find the value of
x
when
z = 1.96,
s = 2.5, and 
Exercise 5.4.56. (Go to Solution)
Solve 
for
S
x
2 ⋅S
x
2
represents a single quantity. Find the value of
S
x
2
when
F = 2.21 and
S
y
2 = 3.24.
Exercise 5.4.57.
Solve 
for
R.
Exercise 5.4.59.
Solve y = 10x + 16 for x.
Exercise 5.4.61.
Solve − 9x + 3y + 15 = 0 for y.
Exercise 5.4.63.
Solve 
for
P.
Exercise 5.4.65.
Solve 
for 
.
Exercises for Review
Solutions to Exercises
5.5. Application I - Translating from Verbal to Mathetical Expressions *
Overview
Translating from Verbal to Mathematical Expressions
Translating from Verbal to Mathematical Expressions
To solve a problem using algebra, we must first express the problem algebraically. To express a problem algebraically, we must scrutinize the wording of the problem to determine the variables and constants that are present and the relationships among them. Then we must translate the verbal phrases and statements to algebraic expressions and equations.
To help us translate verbal expressions to mathematics, we can use the following table as a mathematics dictionary.
| Word or Phrase | Mathematical Operation |
| Sum, sum of, added to, increased by, more than, plus, and | + |
| Difference, minus, subtracted from, decreased by, less, less than | − |
| Product, the product of, of, muitiplied by, times | ⋅ |
| Quotient, divided by, ratio | ÷ |
| Equals, is equal to, is, the result is, becomes | = |
| A number, an unknown quantity, an unknown, a quantity | x (or any symbol) |
Sample Set A
Translate the following phrases or sentences into mathematical expressions or equations.
Example 5.28.
Example 5.29.
Example 5.30.
Example 5.31.
Example 5.32.
Example 5.33.
Practice Set A
Translate the following phrases or sentences into mathematical expressions or equations.
Sometimes the structure of the sentence indicates the use of grouping symbols.
Sample Set B
Translate the following phrases or sentences into mathematical expressions or equations.
Example 5.34.
Commas set off terms.
Example 5.35.
Example 5.36.
Example 5.37.
A number plus one is divided by three times the number minus twelve and the result is four.
Notice that since the phrase "three times the number minus twelve" does not contain a comma, we get the expression
3x − 12
. If the phrase had appeared as "three times the number, minus twelve," the result would have been
Example 5.38.
Some phrases and sentences do not translate directly. We must be careful to read them properly. The word from often appears in such phrases and sentences. The word from means "a point of departure for motion." The following translation will illustrate this use.
The word from indicates the motion (subtraction) is to begin at the point of "some quantity."
Example 5.39.
Eight less than some quantity. Notice that less than could be replaced with from. x − 8
Practice Set B
Translate the following phrases and sentences into mathematical expressions or equations.
Exercise 5.5.10. (Go to Solution)
A number minus five is divided by twice the number plus three and the result is seventeen.
Exercise 5.5.12. (Go to Solution)
An unknown quantity is subtracted from eleven and the result is five less than the unknown quantity.
Exercises
For the following problems, translate the following phrases or sentences into mathematical expressions or equations.
Exercise 5.5.14.
Eight more than a number.
Exercise 5.5.16.
A number minus three.
Exercise 5.5.18.
Negative sixteen minus some quantity.
Exercise 5.5.20.
Ten added to three times some number.
Exercise 5.5.22.
Twice a number is eleven.
Exercise 5.5.24.
One third of a number is two fifths.
Exercise 5.5.26.
Five times a number is that number minus two.
Exercise 5.5.28.
Ten times a number less four results in sixty-six.
Exercise 5.5.30.
Seven more than some number is five more than twice the number.
Exercise 5.5.32.
Eleven fifteenths of two more than a number is eight.
Exercise 5.5.34.
Two more than twice a number is one half the number less three.
Exercise 5.5.36.
Three fifths of a quantity added to the quantity itself is thirty-nine.
Exercise 5.5.37. (Go to Solution)
A number plus seven is divided by two and the result is twenty-two.
Exercise 5.5.38.
Ten times a number minus one is divided by fourteen and the result is one.
Exercise 5.5.39. (Go to Solution)
A number is added to itself then divided by three. This result is then divided by three. The entire result is fifteen.
Exercise 5.5.40.
Ten divided by two more than a number is twenty-one.
Exercise 5.5.42.
Twelve divided by twice a number is fifty-five.
Exercise 5.5.44.
A number divided by itself, plus one, results in seven.
Exercise 5.5.46.
A number plus six, divided by two, is seventy-one.
Exercise 5.5.48.
A number multiplied by itself added to five is thirty-one.
Exercise 5.5.50.
A number is increased by one and then multiplied by five times itself. The result is eighty-four.
Exercise 5.5.51. (Go to Solution)
A number is added to six and that result is multiplied by thirteen. This result is then divided by six times the number. The entire result is equal to fifty-nine.
Exercise 5.5.52.
A number is subtracted from ten and that result is multiplied by four. This result is then divided by three more than the number. The entire result is equal to six.
Exercise 5.5.53. (Go to Solution)
An unknown quantity is decreased by eleven. This result is then divided by fifteen. Now, one is subtracted from this result and five is obtained.
Exercise 5.5.54.
Ten less than some number.
Exercise 5.5.56.
Twelve less than a number.
Exercise 5.5.58.
Sixteen less than some number is forty-two.
Exercise 5.5.60.
Seven is added to ten less than some number. The result is one.
Exercise 5.5.61. (Go to Solution)
Twenty-three is divided by two less than twice some number and the result is thirty-four.
Exercise 5.5.62.
One less than some number is multiplied by three less than five times the number and the entire result is divided by six less than the number. The result is twenty-seven less than eleven times the number.
Exercises for Review
Exercise 5.5.63. (Go to Solution)
(Section 2.3) Supply the missing word. The point on a line that is associated with a particular number is called the __________ of that number.
Exercise 5.5.64.
(Section 2.5) Supply the missing word. An exponent records the number of identical __________ in a multiplication.
Exercise 5.5.65. (Go to Solution)
(Section 3.3) Write the algebraic definition of the absolute value of the number a .
Solutions to Exercises
5.6. Application II - Solving Problems *
Overview
Solving Applied Problems
Solving Applied Problems
Let’s study some interesting problems that involve linear equations in one variable. In order to solve such problems, we apply the following five-step method:
Let x (or some other letter) represent the unknown quantity.
Translate the words to mathematical symbols and form an equation.
Solve this equation.
- Ask yourself "Does this result seem reasonable?" Check the solution by substituting the result into the original statement of the problem.
If the answer doesn’t check, you have either solved the equation incorrectly, or you have developed the wrong equation. Check your method of solution first. If the result does not check, reconsider your equation.
Write the conclusion.
If it has been your experience that word problems are difficult, then follow the five-step method carefully. Most people have difficulty because they neglect step 1.
Always start by INTRODUCING A VARIABLE!
Keep in mind what the variable is representing throughout the problem.
Sample Set A
Example 5.40.
This year an item costs $44 , an increase of $3 over last year’s price. What was last year’s price?
Practice Set A
Exercise 5.6.1. (Go to Solution)
This year an item costs $23 , an increase of $4 over last year’s price. What was last year’s price?
Let x =
Last year's price was __________.
Sample Set B
Example 5.41.
The perimeter (length around) of a square is 60 cm (centimeters). Find the length of a side.
Practice Set B
Exercise 5.6.2. (Go to Solution)
The perimeter of a triangle is 54 inches. If each side has the same length, find the length of a side.
Let x =
The length of a side is __________ inches.
Sample Set C
Example 5.42.
Six percent of a number is 54. What is the number?
Practice Set C
Exercise 5.6.3. (Go to Solution)
Eight percent of a number is 36. What is the number?
Let x =
The number is __________.
Sample Set D
Example 5.43.
An astronomer notices that one star gives off about 3.6 times as much energy as another star. Together the stars give off 55.844 units of energy. How many units of energy does each star emit?
- In this problem we have two unknowns and, therefore, we might think, two variables. However, notice that the energy given off by one star is given in terms of the other star. So, rather than introducing two variables, we introduce only one. The other unknown(s) is expressed in terms of this one. (We might call this quantity the base quantity.)
Let x = number of units of energy given off by the less energetic star. Then, 3.6x = number of units of energy given off by the more energetic star.
Practice Set D
Exercise 5.6.4. (Go to Solution)
Garden A produces 5.8 times as many vegetables as garden B. Together the gardens produce 102 pounds of vegetables. How many pounds of vegetables does garden A produce?
Let x =
Sample Set E
Example 5.44.
Two consecutive even numbers sum to 432. What are the two numbers?
Practice Set E
Exercise 5.6.5. (Go to Solution)
The sum of two consecutive even numbers is 498. What are the two numbers?
Exercises
Solve the following problems. Note that some of the problems may seem to have no practical applications and may not seem very interesting. They, along with the other problems, will, however, help to develop your logic and problem-solving ability.
Exercise 5.6.6. (Go to Solution)
If eighteen is subtracted from some number the result is fifty-two. What is the number?
Let x =
The equation is
(Solve the equation.)
(Check)
The number is __________.
Exercise 5.6.7.
If nine more than twice a number is forty-six, what is the number?
Let x =
The equation is
(Solve the equation.)
(Check)
The number is __________.
Exercise 5.6.8. (Go to Solution)
If nine less than three eighths of a number is two and one fourth, what is the number?
Let x =
The number is __________.
Exercise 5.6.9.
Twenty percent of a number is 68. What is the number?
Let x =
The number is __________.
Exercise 5.6.10. (Go to Solution)
Eight more than a quantity is 37. What is the original quantity?
Let x =
The original quantity is __________.
Exercise 5.6.11.
If a quantity plus 85% more of the quantity is 62.9 , what is the original quantity?
Let x = original quantity.
The original quantity is __________.
Exercise 5.6.12. (Go to Solution)
A company must increase production by 12% over last year’s production. The new output will be 56 items. What was last year’s output?
Let P =
Last year’s output was __________ items.
Exercise 5.6.13.
A company has determined that it must increase production of a certain line of goods by 
times last year’s production. The new output will be 2885 items. What was last year’s output?
Last year’s output was __________ items.
Exercise 5.6.14. (Go to Solution)
A proton is about 1837 times as heavy as an electron. If an electron weighs 2.68 units, how many units does a proton weigh?
A proton weighs __________ units.
Exercise 5.6.15.
Neptune is about 30 times as far from the sun as is the Earth. If it takes light 8 minutes to travel from the sun to the Earth, how many minutes does it take to travel to Neptune?
Light takes __________ minutes to reach Neptune.
Exercise 5.6.16. (Go to Solution)
The radius of the sun is about 695,202 km (kilometers). That is about 109 times as big as the radius of the Earth. What is the radius of the earth?
The radius of the earth is __________ km.
Exercise 5.6.17.
The perimeter of a triangle is 105 cm. If each of the two legs is exactly twice the length of the base, how long is each leg?
Let x = Draw a picture.
Each leg is __________ cm long. The base is __________.
Exercise 5.6.18. (Go to Solution)
A lumber company has contracted to cut boards into two pieces so that one piece is three times the length of the other piece. If a board is 12 feet long, what is the length of each piece after cutting?
The length of the shorter piece is __________ feet, and the length of the longer piece is __________ feet.
Exercise 5.6.19.
A student doing a chemistry experiment has a beaker that contains 84 ml (milliliters) of an alcohol and water solution. Her lab directions tell her that there is 4.6 times as much water as alcohol in the solution. How many milliliters of alcohol are in the solution? How many milliliters of water?
There are __________ ml of alcohol in the solution. There are __________ ml of water in the solution.
Exercise 5.6.20. (Go to Solution)
A statistician is collecting data to help him estimate the average income of accountants in California. He needs to collect 390 pieces of data and he is 
done. How many pieces of data has the statistician collected?
The statistician has collected __________ pieces of data.Suppose the statistician is 4 pieces of data short of being
done. How many pieces of data has he collected?
Exercise 5.6.21.
A television commercial advertises that a certain type of battery will last, on the average, 20 hours longer than twice the life of another type of battery. If consumer tests show that the advertised battery lasts 725 hours, how many hours must the other type of battery last for the advertiser’s claim to be valid?
The other type of battery must last __________ hours for the advertiser’s claim to be valid.
Exercise 5.6.22. (Go to Solution)
A 1000-ml flask containing a chloride solution will fill 3 beakers of the same size with 210 ml of the solution left over. How many milliliters of the chloride solution will each beaker hold?
Each beaker will hold __________ ml of the chloride solution.
Exercise 5.6.23.
A star burns 
of its original mass then blows off 
of the remaining mass as a planetary nebula. If the final mass is 3 units of mass, what was the original mass?
The original mass was __________ units of mass.
Exercise 5.6.25.
When eleven is subtracted from a number, the result is 85. What is the number?
Exercise 5.6.26. (Go to Solution)
Three times a number is divided by 6 and the result is 10.5 . What is the number?
Exercise 5.6.27.
When a number is multiplied by itself, the result is 144. What is the number?
Exercise 5.6.28. (Go to Solution)
A number is tripled, then increased by seven. The result is 48. What is the number?
Exercise 5.6.29.
Eight times a number is decreased by three times the number, giving a difference of 22. What is the number?
Exercise 5.6.30. (Go to Solution)
One number is fifteen more than another number. The sum of the two numbers is 27. What are they?
Exercise 5.6.31.
The length of a rectangle is 6 meters more than three times the width. The perimeter of the rectangle is 44 meters What are the dimensions of the rectangle?
Exercise 5.6.32. (Go to Solution)
Seven is added to the product of 41 and some number. The result, when divided by four, is 63. What is the number?
Exercise 5.6.33.
The second side of a triangle is five times the length of the smallest side. The third is twice the length of the second side. The perimeter of the triangle is 48 inches. Find the length of each side.
Exercise 5.6.34. (Go to Solution)
Person A is four times as old as person B, who is six times as old as person C, who is twice as old as person D. How old is each person if their combined ages are 189 months?
Exercise 5.6.35.
Two consecutive odd integers sum to 151. What are they?
Exercise 5.6.37.
Three consecutive even integers add up to 131. What are they?
Exercise 5.6.38. (Go to Solution)
As a consequence of Einstein’s theory of relativity, the rate of time passage is different for a person in a stationary position and a person in motion. (Hard to believe, but true!) To the moving observer, the rate of time passage is slower than that of the stationary observer, that is, the moving person ages slower than the stationary observer. (This fact has been proven many times by experiments with radioactive materials.) The effect is called “time dilation” and is really only noticeable when an object is traveling at near the speed of light (186,000 miles per second). Considering these ideas, try to solve the following problems:
Two people have identical clocks. One is standing on the earth and the other is moving in a spacecraft at 95% the speed of light, 176,700 miles per second. The moving person’s rate of time passage at this speed is about 0.31 times as fast as the person standing on earth.
If two days of earth time pass, how many days actually pass on the spacecraft?
If 30 years of earth time pass, how many years have actually passed on the spacecraft?
__________ years have passed on the spacecraft.
If 30 years have passed on the spacecraft, how many years have passed on the earth?
A space traveler makes a round-trip voyage to the star Capella. The trip takes her 120 years (traveling at 176,000 miles per second). If it is the year 2000 on earth when she leaves, what earth year will it be when she returns?
Exercises for Review
Exercise 5.6.40. (Go to Solution)
(Section 5.2) Classify the equation x + 4 = 1 as an identity, a contradiction, or a conditional equation.
Exercise 5.6.41.
(Section 5.2) Classify the equation 2x + 3 = 2x + 3 as an identity, a contradiction or a conditional equation.
Exercise 5.6.43.
(Section 5.5) Translate the following sentence to a mathematical equation. Three less than an unknown number is multiplied by negative four. The result is two more than the original unknown number.
Solutions to Exercises
Solution to Exercise 5.6.6. (Return to Exercise)
Step 1: Let x = the unknown quantity.Step 2: The equation is x − 18 = 52. Step 3: (Solve the equation.) Add 18 to each side. x − 18 + 18 = 52 + 18 x = 70 Step 4: (Check) 70 − 18 = 52; True.Step 5: The number is 70.
Solution to Exercise 5.6.18. (Return to Exercise)
Step 5: The length of the shorter piece is 3 feet, and the length of the longer piece is 9 feet.
Solution to Exercise 5.6.20. (Return to Exercise)
Step 5: The statistician has collected 260 pieces of data.
Solution to Exercise 5.6.22. (Return to Exercise)
Step 5: Each beaker will hold 
ml of chloride solution.
Solution to Exercise 5.6.34. (Return to Exercise)
Step 5: The age of D is 3 months; C is 6 months; B is 36 months; A is 144 months.
Solution to Exercise 5.6.36. (Return to Exercise)
Step 5: The first integer is 11; second is 12; third is 13.
Solution to Exercise 5.6.38. (Return to Exercise)
(a) Step 5: The time passed in space is 0.62 days.(b) Step 5: 9.3 years have passed on the spacecraft. (c) Step 5: 96.77 years have passed on the earth.(d) Step 5: Earth year when she returns will be 2387.
5.7. Linear inequalities in One Variable *
Overview
Inequalities
Linear Inequalities
The Algebra of Linear Inequalities
Compound Inequalities
Inequalities
We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. Not all relationships need be relationships of equality, however. Certainly the number of human beings on earth is greater than 20. Also, the average American consumes less than 10 grams of vitamin C every day. These types of relationships are not relationships of equality, but rather, relationships of inequality.
Linear Inequalities
A linear inequality is a mathematical statement that one linear expression is greater than or less than another linear expression.
The following notation is used to express relationships of inequality:
Note that the expression x > 12 has infinitely many solutions. Any number strictly greater than 12 will satisfy the statement. Some solutions are 13, 15, 90, 12.1, 16.3 and 102.51 .
Sample Set A
The following are linear inequalities in one variable.
Example 5.45.
x ≤ 12
x + 7 > 4
y + 3 ≥ 2y − 7
P + 26 < 10(4P − 6)
The following are not linear inequalities in one variable.
Example 5.46.
x 2 < 4 .The term x 2 is quadratic, not linear.
x ≤ 5y + 3 .There are two variables. This is a linear inequality in two variables.
y + 1 ≠ 5 .Although the symbol ≠ certainly expresses an inequality, it is customary to use only the symbols < , > , ≤ , ≥ .
Practice Set A
A linear equation, we know, may have exactly one solution, infinitely many solutions, or no solution. Speculate on the number of solutions of a linear inequality. (Hint: Consider the inequalities x < x − 6 and x ≥ 9 .)
A linear inequality may have infinitely many solutions, or no solutions.
The Algebra of Linear Inequalities
Inequalities can be solved by basically the same methods as linear equations. There is one important exception that we will discuss in item 3 of the algebra of linear inequalities.
Let
a, b, and c
represent real numbers and assume that
Then,
if a < b,
.If any real number is added to or subtracted from both sides of an inequality, the sense of the inequality remains unchanged.
If c is a positive real number, then if a < b,
If both sides of an inequality are multiplied or divided by the same positive number the sense of the inequality remains unchanged.
If c is a negative real number, then if a < b,
If both sides of an inequality are multiplied or divided by the same negative number, the inequality sign must be reversed (change direction) in order for the resulting inequality to be equivalent to the original inequality. (See problem 4 in the next set of examples.)
For example, consider the inequality 3 < 7 .
Example 5.47.
For 3 < 7 , if 8 is added to both sides, we get
Example 5.48.
For 3 < 7 , if 8 is subtracted from both sides, we get
Example 5.49.
For 3 < 7 , if both sides are multiplied by 8 (a positive number), we get
Example 5.50.
For 3 < 7 , if both sides are multiplied by − 8 (a negative number), we get
( − 8)3 > ( − 8)7
Notice the change in direction of the inequality sign.
If we had forgotten to reverse the direction of the inequality sign we would have obtained the incorrect statement − 24 < − 56 .
Example 5.51.
For 3 < 7 , if both sides are divided by 8 (a positive number), we get
Example 5.52.
For 3 < 7 , if both sides are divided by − 8 (a negative number), we get
Sample Set B
Solve the following linear inequalities. Draw a number line and place a point at each solution.
Example 5.53.
Thus, all numbers strictly greater than 5 are solutions to the inequality
3x > 15
. 
Example 5.54.
Example 5.55.
Example 5.56.
Example 5.57.
Practice Set B
Solve the following linear inequalities.
Compound Inequalities
Another type of inequality is the compound inequality. A compound inequality is of the form:
a < x < b
There are actually two statements here. The first statement is a < x . The next statement is x < b . When we read this statement we say " a is less than x ," then continue saying "and x is less than b ."
Just by looking at the inequality we can see that the number x is between the numbers a and b . The compound inequality a < x < b indicates "betweenness." Without changing the meaning, the statement a < x can be read x > a . (Surely, if the number a is less than the number x , the number x must be greater than the number a .) Thus, we can read a < x < b as " x is greater than a and at the same time is less than b ." For example:
4 < x < 9 .The letter x is some number strictly between 4 and 9. Hence, x is greater than 4 and, at the same time, less than 9. The numbers 4 and 9 are not included so we use open circles at these points.
− 2 < z < 0 .The z stands for some number between − 2 and 0. Hence, z is greater than − 2 but also less than 0.
1 < x + 6 < 8 .The expression x + 6 represents some number strictly between 1 and 8. Hence, x + 6 represents some number strictly greater than 1, but less than 8.
.The term 
represents some number between and including 
and 
. Hence, 
represents some number greater than or equal to 
to but less than or equal to 
.
Consider problem 3 above, 1 < x + 6 < 8 . The statement says that the quantity x + 6 is between 1 and 8. This statement will be true for only certain values of x . For example, if x = 1 , the statement is true since 1 < 1 + 6 < 8 . However, if x = 4.9 , the statement is false since 1 < 4.9 + 6 < 8 is clearly not true. The first of the inequalities is satisfied since 1 is less than 10.9 , but the second inequality is not satisfied since 10.9 is not less than 8.
We would like to know for exactly which values of x the statement 1 < x + 6 < 8 is true. We proceed by using the properties discussed earlier in this section, but now we must apply the rules to all three parts rather than just the two parts in a regular inequality.
Sample Set C
Example 5.58.
Solve 1 < x + 6 < 8 .
Thus, if x is any number strictly between − 5 and 2, the statement 1 < x + 6 < 8 will be true.
Example 5.59.
Solve 
.
Thus, if
x
is any number between 
and 4, the original inequality will be satisfied.
Practice Set C
Find the values of x that satisfy the given continued inequality.
Exercises
For the following problems, solve the inequalities.
Exercise 5.7.18.
y − 5 ≤ 8
Exercise 5.7.20.
x − 5 > 16
Exercise 5.7.22.
9y − 12 ≤ 6
Exercise 5.7.24.
4x − 14 > 21
Exercise 5.7.26.
− 8x < 40
Exercise 5.7.28.
− 3y > 39
Exercise 5.7.30.
Exercise 5.7.32.
Exercise 5.7.34.
Exercise 5.7.36.
Exercise 5.7.38.
Exercise 5.7.40.
− 3x + 7 ≤ − 5
Exercise 5.7.42.
6x − 11 < 31
Exercise 5.7.44.
Exercise 5.7.46.
4(x + 1) > − 12
Exercise 5.7.48.
3( − x + 3) > − 27
Exercise 5.7.50.
− 7(x − 77) ≤ 0
Exercise 5.7.52.
6y + 12 ≤ 5y − 1
Exercise 5.7.54.
4x + 5 > 5x − 11
Exercise 5.7.56.
− 2x − 7 > 5x
Exercise 5.7.58.
3 − x ≥ 4
Exercise 5.7.60.
2 − 4x ≤ − 3 + x
Exercise 5.7.62.
2[6 + 2(3x − 7)] ≥ 4
Exercise 5.7.64.
− 2(4x − 1) < 3(5x + 8)
Exercise 5.7.66.
− .0091x ≥ 2.885x − 12.014
Exercise 5.7.67. (Go to Solution)
What numbers satisfy the condition: twice a number plus one is greater than negative three?
Exercise 5.7.68.
What numbers satisfy the condition: eight more than three times a number is less than or equal to fourteen?
Exercise 5.7.69. (Go to Solution)
One number is five times larger than another number. The difference between these two numbers is less than twenty-four. What are the largest possible values for the two numbers? Is there a smallest possible value for either number?
Exercise 5.7.70.
The area of a rectangle is found by multiplying the length of the rectangle by the width of the rectangle. If the length of a rectangle is 8 feet, what is the largest possible measure for the width if it must be an integer (positive whole number) and the area must be less than 48 square feet?
Exercises for Review
Exercise 5.7.75. (Go to Solution)
(Section 5.6) The perimeter of a triangle is 40 inches. If the length of each of the two legs is exactly twice the length of the base, how long is each leg?
Solutions to Exercises
Solution to Exercise 5.7.69. (Return to Exercise)
First number: any number strictly smaller that 6.Second number: any number strictly smaller than 30. No smallest possible value for either number.No largest possible value for either number.
5.8. Linear Equations in Two Variables *
Overview
Solutions to Linear Equations in Two Variables
Ordered Pairs as Solutions
Solutions to Linear Equations in Two Variables
We have discovered that an equation is a mathematical way of expressing the relationship of equality between quantities. If the relationship is between two quantities, the equation will contain two variables. We say that an equation in two variables has a solution if an ordered pair of values can be found such that when these two values are substituted into the equation a true statement results. This is illustrated when we observe some solutions to the equation y = 2x + 5 .
x = 4, y = 13; since 13 = 2(4) + 5 is true .
x = 1, y = 7; since 7 = 2(1) + 5 is true .
x = 0, y = 5; since 5 = 2(0) + 5 is true .
x = − 6, y = − 7; since -7 = 2( − 6) + 5 is true .
Ordered Pairs as Solutions
It is important to keep in mind that a solution to a linear equation in two variables is an ordered pair of values, one value for each variable. A solution is not completely known until the values of both variables are specified.
Recall that, in an equation, any variable whose value can be freely assigned is said to be an independent variable. Any variable whose value is determined once the other value or values have been assigned is said to be a dependent variable. If, in a linear equation, the independent variable is x and the dependent variable is y , and a solution to the equation is x = a and y = b , the solution is written as the
ORDERED PAIR (a, b)
In an ordered pair, (a, b) , the first component, a , gives the value of the independent variable, and the second component, b , gives the value of the dependent variable.
We can use ordered pairs to show some solutions to the equation y = 6x − 7 .
Example 5.60.
(0, − 7) .If x = 0 and y = − 7 , we get a true statement upon substitution and computataion.
Example 5.61.
(8, 41) .If x = 8 and y = 41 , we get a true statement upon substitution and computataion.
Example 5.62.
( − 4, − 31) .If x = − 4 and y = − 31 , we get a true statement upon substitution and computataion.
These are only three of the infintely many solutions to this equation.
Sample Set A
Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.
Example 5.63.
y = 3x − 6, if x = 1
Substitute 1 for x , compute, and solve for y .
Hence, one solution is (1, − 3) .
Example 5.64.
y = 15 − 4x, if x = − 10
Substitute − 10 for x , compute, and solve for y .
Hence, one solution is ( − 10, 55) .
Example 5.65.
b = − 9a + 21, if a = 2
Substitute 2 for a , compute, and solve for b .
Hence, one solution is (2, 3) .
Example 5.66.
5x − 2y = 1, if x = 0
Substitute 0 for x , compute, and solve for y .
Hence, one solution is 
.
Practice Set A
Find a solution to each of the following linear equations in two variables and write the solution as an ordered pair.
Exercises
For the following problems, solve the linear equations in two variables.
Exercise 5.8.7.
y = − 2x + 1, if x = 0
Exercise 5.8.9.
x + y = 7, if x = 8
Exercise 5.8.11.
Exercise 5.8.13.
− 4x − 4y = 4, if y = 7
Exercise 5.8.15.
Exercise 5.8.17.
x + y = 0, if x = 0
Exercise 5.8.19.
y + 17 = x, if x = − 12
Exercise 5.8.21.
Exercise 5.8.23.
y + 7 − x = 0, if x =
Exercise 5.8.25.
436x + 189y = 881, if x = − 4231
Exercise 5.8.27.
y = 2(4x + 5), if x = − 1
Exercise 5.8.29.
3y = 4(4x + 1), if x = − 3
Exercise 5.8.31.
− 8y = 7(8x + 2), if x = 0
Exercise 5.8.33.
b = − 5a + 21, if a = − 9
Exercise 5.8.35.
− 5m + 11 = n + 1, if n = 4
Exercise 5.8.37.
7(t − 6) = 10(2 − s), if s = 5
Exercise 5.8.39.
2y = 0x − 11, if x = − 7
Exercise 5.8.41.
− 5y = 0x − 1, if x = 0
Exercise 5.8.43.
y = 0(3x + 9) − 1, if x = 12
Calculator Problems
Exercise 5.8.44. (Go to Solution)
An examination of the winning speeds in the Indianapolis 500 automobile race from 1961 to 1970 produces the equation y = 1.93x + 137.60 , where x is the number of years from 1960 and y is the winning speed. Statistical methods were used to obtain the equation, and, for a given year, the equation gives only the approximate winning speed. Use the equation y = 1.93x + 137.60 to find the approximate winning speed in
1965
1970
1986
1990
Exercise 5.8.45.
In electricity theory, Ohm’s law relates electrical current to voltage by the equation y = 0.00082x , where x is the voltage in volts and y is the current in amperes. This equation was found by statistical methods and for a given voltage yields only an approximate value for the current. Use the equation y = 0.00082x to find the approximate current for a voltage of
6 volts
10 volts
Exercise 5.8.46. (Go to Solution)
Statistical methods have been used to obtain a relationship between the actual and reported number of German submarines sunk each month by the U.S. Navy in World War II. The equation expressing the approximate number of actual sinkings, y , for a given number of reported sinkings, x , is y = 1.04x + 0.76 . Find the approximate number of actual sinkings of German submarines if the reported number of sinkings is
4
9
10
Exercise 5.8.47.
Statistical methods have been used to obtain a relationship between the heart weight (in milligrams) and the body weight (in milligrams) of 10-month-old diabetic offspring of crossbred male mice. The equation expressing the approximate body weight for a given heart weight is y = 0.213x – 4.44 . Find the approximate body weight for a heart weight of
210 mg
245 mg
Exercise 5.8.48. (Go to Solution)
Statistical methods have been used to produce the equation y = 0.176x − 0.64 . This equation gives the approximate red blood cell count (in millions) of a dog’s blood, y , for a given packed cell volume (in millimeters), x . Find the approximate red blood cell count for a packed cell volume of
40 mm
42 mm
Exercise 5.8.49.
An industrial machine can run at different speeds. The machine also produces defective items, and the number of defective items it produces appears to be related to the speed at which the machine is running. Statistical methods found that the equation y = 0.73x − 0.86 is able to give the approximate number of defective items, y , for a given machine speed, x . Use this equation to find the approximate number of defective items for a machine speed of
9
12
Exercise 5.8.50. (Go to Solution)
A computer company has found, using statistical techniques, that there is a relationship between the aptitude test scores of assembly line workers and their productivity. Using data accumulated over a period of time, the equation y = 0.89x − 41.78 was derived. The x represents an aptitude test score and y the approximate corresponding number of items assembled per hour. Estimate the number of items produced by a worker with an aptitude score of
80
95
Exercise 5.8.51.
Chemists, making use of statistical techniques, have been able to express the approximate weight of potassium bromide, W , that will dissolve in 100 grams of water at T degrees centigrade. The equation expressing this relationship is W = 0.52T + 54.2 . Use this equation to predict the potassium bromide weight that will dissolve in 100 grams of water that is heated to a temperature of
70 degrees centigrade
95 degrees centigrade
Exercise 5.8.52. (Go to Solution)
The marketing department at a large company has been able to express the relationship between the demand for a product and its price by using statistical techniques. The department found, by analyzing studies done in six different market areas, that the equation giving the approximate demand for a product (in thousands of units) for a particular price (in cents) is y = − 14.15x + 257.11 . Find the approximate number of units demanded when the price is
$0.12
$0.15
Exercise 5.8.53.
The management of a speed-reading program claims that the approximate speed gain (in words per minute), G , is related to the number of weeks spent in its program, W , is given by the equation G = 26.68W − 7.44 . Predict the approximate speed gain for a student who has spent
3 weeks in the program
10 weeks in the program
Exercises for Review
Solutions to Exercises
Solution to Exercise 5.8.44. (Return to Exercise)
(a) Approximately 147 mph using ( 5,147.25 ) (b) Approximately 157 mph using ( 10,156.9 ) (c) Approximately 188 mph using ( 26,187.78 ) (d) Approximately 196 mph using ( 30,195.5 )
Solution to Exercise 5.8.46. (Return to Exercise)
(a) Approximately 5 sinkings using ( 4,4.92 ) (b) Approximately 10 sinkings using ( 9,10.12 ) (c) Approximately 11 sinkings using ( 10,11.16 )
Solution to Exercise 5.8.48. (Return to Exercise)
(a) Approximately 6.4 using ( 40,6.4 ) (b) Approximately 4.752 using ( 42,7.752 )
Solution to Exercise 5.8.50. (Return to Exercise)
(a) Approximately 29 items using ( 80,29.42 ) (b) Approximately 43 items using ( 95,42.77 )
Solution to Exercise 5.8.52. (Return to Exercise)
(a) Approximately 87 units using ( 12,87.31 ) (b) Approximately 45 units using ( 15,44.86 )
5.9. Summary of Key Concepts*
Summary of Key Concepts
An equation that is true for all acceptable values of the variable is called identity. x + 3 = x + 3 is an identity.
Contradictions are equations that are never true regardless of the value substituted for the variable. x + 1 = x is a contradiction.
An equation whose truth is conditional upon the value selected for the variable is called a conditional equation.
The collection of values that make an equation true are called the solutions of the equation. An equation is said to be solved when all its solutions have been found.
Equations that have precisely the same collection of solutions are called equivalent equations.An equivalent equation can be obtained from a particular equation by applying the same binary operation to both sides of the equation, that is,
adding or subtracting the same number to or from both sides of that particular equation.
multiplying or dividing both sides of that particular equation by the same non-zero number.
A literal equation is an equation that is composed of more than one variable.
If, when solving an equation, all the variables are eliminated and a true statement results, the equation is an identity.
If, when solving an equation, all the variables are eliminated and a false statement results, the equation is a contradiction.
When solving word problems it is absolutely necessary to know how certain words translate into mathematical symbols.
Let x (or some other letter) represent the unknown quantity.
Translate the words to mathematics and form an equation. A diagram may be helpful.
Solve the equation.
Check the solution by substituting the result into the original statement of the problem.
Write a conclusion.
A linear inequality is a mathematical statement that one linear expression is greater than or less than another linear expression.
An inequality of the form a < x < b is called a compound inequality.
A pair of values that when substituted into an equation in two variables produces a true statement is called a solution to the equation in two variables. These values are commonly written as an ordered pair. The expression (a, b) is an ordered pair. In an ordered pair, the independent variable is written first and the dependent variable is written second.
5.10. Exercise Supplement *
Exercise Supplement
Solving Equations (Section 5.2) - Further Techniques in Equation Solving (Section 5.4)
Solve the equations for the following problems.
Exercise 5.10.2.
a − 7 = 4
Exercise 5.10.4.
a + 2 = 0
Exercise 5.10.6.
x − 5 = − 6
Exercise 5.10.8.
y − 4 = 4
Exercise 5.10.10.
4x = 24
Exercise 5.10.12.
6m = − 30
Exercise 5.10.14.
− 8y = − 72
Exercise 5.10.16.
− y = − 10
Exercise 5.10.18.
6x − 1 = 29
Exercise 5.10.20.
6x − 5 = − 29
Exercise 5.10.22.
9a + 5 = − 22
Exercise 5.10.24.
Exercise 5.10.26.
Exercise 5.10.28.
Exercise 5.10.30.
Exercise 5.10.32.
Exercise 5.10.34.
Exercise 5.10.36.
Exercise 5.10.38.
Exercise 5.10.40.
Exercise 5.10.42.
− 2(a − 3) = 16
Exercise 5.10.44.
3x + 7 = 5x − 21
Exercise 5.10.46.
Solve I = p r t for t. Find the value of t when I = 3500, P = 3000, and r = 0.05.
Exercise 5.10.48.
Solve p = m v for m. Find the value of m when p = 4240 and v = 260.
Exercise 5.10.49. (Go to Solution)
Solve P = R − C for R. Find the value of R when P = 480 and C = 210.
Exercise 5.10.50.
Exercise 5.10.52.
Solve 3y − 6x = 12 for y.
Exercise 5.10.54.
Application I - Translating from Verbal to Mathetical Expressions (Section 5.5)
For the following problems, translate the phrases or sentences to mathematical expressions or equations.
Exercise 5.10.56.
A quantity less eight.
Exercise 5.10.58.
Negative ten minus some number.
Exercise 5.10.60.
One seventh of a number plus two ninths of the number.
Exercise 5.10.62.
Twice a quantity plus nine is equal to the quantity plus sixty.
Exercise 5.10.63. (Go to Solution)
Four times a number minus five is divided by seven. The result is ten more than the number.
Exercise 5.10.64.
A number is added to itself five times, and that result is multiplied by eight. The entire result is twelve.
Exercise 5.10.66.
A quantity less three is divided by two more than the quantity itself. The result is one less than the original quantity.
Exercise 5.10.67. (Go to Solution)
A number is divided by twice the number, and eight times the number is added to that result. The result is negative one.
Exercise 5.10.68.
An unknown quantity is decreased by six. This result is then divided by twenty. Ten is subtracted from this result and negative two is obtained.
Exercise 5.10.69. (Go to Solution)
One less than some number is divided by five times the number. The result is the cube of the number.
Exercise 5.10.70.
Nine less than some number is multiplied by the number less nine. The result is the square of six times the number.
Application II - Solving Problems (Section 5.6)
For the following problems, find the solution.
Exercise 5.10.71. (Go to Solution)
This year an item costs $106, an increase of $10 over last year’s price. What was last year’s price?
Exercise 5.10.72.
The perimeter of a square is 44 inches. Find the length of a side.
Exercise 5.10.74.
Two consecutive integers sum to 63. What are they?
Exercise 5.10.76.
If twenty-one is subtracted from some number and that result is multiplied by two, the result is thirty-eight. What is the number?
Exercise 5.10.78.
A statistician is collecting data to help her estimate the number of pickpockets in a certain city. She needs 108 pieces of data and is 
done. How many pieces of data has she collected?
Exercise 5.10.79. (Go to Solution)
The statistician in problem 78 is eight pieces of data short of being 
done. How many pieces of data has she collected?
Exercise 5.10.80.
A television commercial advertises that a certain type of light bulb will last, on the average, 200 hours longer than three times the life of another type of bulb. If consumer tests show that the advertised bulb lasts 4700 hours, how many hours must the other type of bulb last for the advertiser’s claim to be valid?
Linear inequalities in One Variable (Section 5.7)
Solve the inequalities for the following problems.
Exercise 5.10.82.
x − 6 ≥ 12
Exercise 5.10.84.
5x − 14 < 1
Exercise 5.10.86.
− 2y ≥ 14
Exercise 5.10.88.
Exercise 5.10.90.
Exercise 5.10.92.
− 4c + 3 ≤ 5
Exercise 5.10.94.
3(4x − 5) > − 6
Exercise 5.10.96.
5x + 4 ≥ 7x + 16
Exercise 5.10.98.
4(6x + 1) + 2 ≥ − 3(x − 1) + 4
Exercise 5.10.100.
What numbers satisfy the condition: nine less than negative four times a number is strictly greater than negative one?
Linear Equations in Two Variables (Section 5.8)
Solve the equations for the following problems.
Exercise 5.10.102.
y = − 10x + 11, if x = − 1
Exercise 5.10.104.
4m + 2k = 30, if m = 8
Exercise 5.10.106.
y = − 2(7x − 4), if x = − 1
Exercise 5.10.108.
6(t + 8) = − (a − 5), if a = 10
Exercise 5.10.110.
− a(a + 1) = 2b + 1, if a = − 2
Solutions to Exercises
5.11. Proficiency Exam *
Proficiency Exam
Solve the equations and inequalities for the following problems.
Translate the phrases or sentences into mathematical expressions or equations for the following problems.
Exercise 5.11.20. (Go to Solution)
(Section 5.5) A number is added to itself and this result is multiplied by the original number cubed. The result is twelve.
Exercise 5.11.21. (Go to Solution)
(Section 5.5) A number is decreased by five and that result is divided by ten more than the original number. The result is six times the original number.
Solve the following problems.
Exercise 5.11.22. (Go to Solution)
(Section 5.6) Eight percent of a number is 1.2 . What is the number?
Exercise 5.11.23. (Go to Solution)
(Section 5.6) Three consecutive odd integers sum to 38. What are they?
Exercise 5.11.24. (Go to Solution)
(Section 5.6) Five more than three times a number is strictly less than seventeen. What is the number?
Exercise 5.11.25. (Go to Solution)
(Section 5.8) Solve y = 8x − 11 for y if x = 3 , and write the solution as an ordered pair.
Solutions to Exercises
Solution to Exercise 5.11.23. (Return to Exercise)
There are no three consecutive odd integers that add to 38.